I found this section of PT56 to be very difficult. In retrospect, I see it really played to my weaknesses. So to get over it, I thought I would figure them out and act as if I was teaching it to someone else. And as long as I was doing that, I figured I would post it on here.

So on to question one, which took a stupid 13 minutes for me to do.

This is a pretty straightforward linear game with only one level to worry about. The problem is that the rules, although they contain overlapping elements, are not easily combined into a single diagram. I've seen other people use four templates to solve this, but this is not something that I think I would do on a test day. It just seems too risky, in that you could go through all the work and one error messes you up, or perhaps it's even not helpful.

It basically just has the not laws inferred from the rules, and the G < K rule is combined with the rule that G and L are next to each other with a G/L < K rule. Note F is a floater

On to the questions!

1. This is your basic apply a rule question, nothing too difficult, E turns out to be the right answer

2. This is a global question, kind of scary to see on an already scary first game. But if you just take them one at a time:

A seems very plausible and since J and L are only separated by one it seems even likely that L will also be before H. I didn't cross it off but moved on. Then B just follows from the basic G/L<K inference, so it's the right answer. Quite a relief.

3. So this question would seem to follow straight from our not-laws in the diagram. But given the lack of a cohesive diagram and the possibility of weird interactions, I decided to just make a quick side diagram to see if K could indeed be 3rd:

G-L-K-J-H/F-F/H

Yup, and it turned out this diagram would come in handy later!

4. This was the hardest one to this point. There are basically two options here, start plugging in like mad or try to solve it theoretically. I just plugged in, but in retrospect there was a better way. I think the best thing to do is think about which application would cause the greatest amount of restrictions. I liked the ones that applied J or L to 2 or 5 o'clock because these led to a clear placing of the other L/J and blocked off a space. Then the J at 5 is especially nice because it pins in the H as well, a double force.

Plugging it in shows this is the right answer.

5. This one seemed damned intimidating at first, but I think this is one where it plays to go with your instincts. I didn't and tried something that seemed more appropriate. But here is what I think is the right way to do it.

If F can't be first, then only G/J/L can be. Although it may seem like a bad idea, go ahead and do a minidiagram for all of those. They each allow some placement because each are involved in big rules.

then looking at the answer choices, A seems imminently possible. the others seem very impossible given these were used up in all the choices. Should have gone with my gut!

6. I messed this one up big time. E, in retrospect, follows from my initial not laws. I figured I should go through all of them and this took tons of time.

an underfunded game such that variables would have to be used multiple times. The distributions would be either 3-1-1-1 or 2-2-1-1.

No big inferences here that I saw. Three rules that don't interact in any obviously interesting ways. I did note that the recliner was mentioned in two of them, so that it could easily get filled up.

The numerical distribution didn't seem that helpful. I guess it's nice to know that, at least, two of them would only move one item.

On to the questions:

7. We've all seen this before. Just apply the rules starting with the GJ not block; then the double conditional; then the regular conditional. You're left with A. easy point.

8. I made a small side diagram here. Right away the inference is that R is full so anything that forces something else into R can't happen. This pops up with the double conditional Gs <--> Hr. So we know G can't be in S, so it's in T and this is one of the choices (D) bingo!

9. Here's another easy one, and one where the numerical distribution comes into play. Heather does everything so you know the other three only do one each. Also because of the double conditional G does the sofa. this is all you need to knowA. G can't do R if she does S B. Credited responseC. J can't move sofa because G is doing itD. dittoE. G can't do R if she does S

10. I think the best thing here is to realize there are only 4 movers so use that to know that whichever pair we are talking about moving the R and T, the other two will have to do S.

A. GJ is explicitly ruled out by the last ruleB. CorrectC. if J is T them M is R and that can't be.D. forces Gj in the middle which can't be - this slowed me for a bit. good to keep these rules in your head if you canE. this is the best distractor I think. It gets you thinking it's right because you know that Jt forces Mr, which this answer would seem to lead from . But think that GH is left in the middle, which would cause a contradiction since H should be forced to R if G is Sofa.

11. I love local games. side diagram it. H is forced into the table because she can't do R because that would force G into a full sofa. That's all you need to solve it! Sometimes it pays to just move into the answer choices even if it seems like you can do more!

A. nope heather is only doing the tableB. sameC. GJ is impossible, for the umpteenth time!d. impossible since heather is thereE. correct!

Okay on to group three. This was a pretty easy one where you can make up so much needed time. The sad thing, this took me longer and I did worse the second time I did it, making me wonder if the studying I have done has been worth anything at all.

I realized later on that this is perfect for templates, yielding three nearly complete templates that make solving it an absolute breeze.

I actually came about this somewhat differently the first time:

-Pretty simple set up, the two parks with three open spots. You know you will use 3/4 trees in each of the parks. This means quite a deal of overlap (2 trees). The numerical distribution is 2-2-1-1. You can't use any tree three times because then they would double up on a park, which is not allowed.

Now originally I didn't use templates but saw this contrapositive and realized that if T is not there, then O is not there, which is impossible because then only 2 trees can be placed. So T always has to be there. This was enough to get me through, but realizing that templates are created would have been more efficient. You have to think in these cases, do any of the rules produce a limited number of overall patterns? Here the MS @ least once does and leads to three kick ass templates. (see diagram)

With these the questions are dead easy. Many can probably do this in 5 minutes or maybe even less.

12. Use templates or rule application to whittle down to the correct answer: D

13. T has to be in each park, so C is correct.

14. Refer to the first and third template and trust your stuff. Right out of the gate A is possible

15. Again refer to templates and see that A cannot be true.

16. Use templates to find the right answer. This one takes some mental gymnastics but hey it's the LSAT right buddy? We are looking for what could be true. 4 will be impossibleA. this was basically ruled out by the last one. We know T is in both and we know that O can't be in both. cross it offB. Intuitively ruled out by 15 as well. We need at least one S and can have no more than one O. So no way we have more Os than SsC. Ruled out by my initial inference from my kick ass contrapositive. You're welcome!D. Obviously not trueE. If you trust your stuff just circle this one and go. Or recognize that it's describing the third template.

Summary: Great game. I think it gives people problems because it is on the heels of very different games so you have to change your though processes. Lsat authors love to do this kind of thing. It might make sense to just close your eyes for 5 secs between each game and go to your happy place.

Ok here is the last set (the executive game 17-23) from PT56. This one was shit for me but I think that was because I was mind-effed from the prior ones. This one is actually very doable with standard techniques. I've seen some setups online but I think mine is actually superior (that's a first).

We have five Execs inspecting three factories in a set order. It's an advanced linear game with three components. It is a balanced game with 5 variables and 5 slots for them, although we do not know know exactly how many will be assigned to each factory (from 1-3). Numerical distribution will be important

Numerical Distribution: Rule 2 tells us Farmington (did anyone else start thinking about the shield?) will have one exec inspect it. So this means that the others will have to take up the slack: 2-2 or 3-1, although we don't know which one. You can see this in my setup image.

we have F < H which tell us F will be one of the first two days and H will be one of the second two.

Q < R and T, which tells us Q can't be last and R or T can't be first

and S <= V which doesn't lead to any not laws, but is good to keep in mind.

so plenty of not laws, no big inferences, but using this setup and knowledge that if we figure out which factories go where and which execs are assigned where, we can get an idea of the numbers and make some inferences pretty easily. Go with confidence!

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17. This is a grab a rule that is totally doable just by taking the rules in order of ease of application

18. Made a side diagram. Easier said that done, I know, since this damn one gives so little room. I honestly think that room allowed is one of the major factors determining performance.

So put RT in slot 2. We know that slot two is not the Farm (because it already has 2 people) so that means slot one is, and it can only hold one. This is Q because Q is before RT.

Check the question, it is a Must Be True. And the very first choice gives it to us. A: Quinn is in the Farm, as we know

19. This one is another example of my fundamental rules. Sometimes, even if you think you haven't made all the inferences you can, head to the answer choices because that is often all you need.

Also, making a makeshift diagram is better than freaking out that you can't make a perfect one, and thus wasting time.

I started by just placing Q & S vertically. Then saying they must be in front of R and T because of the Q < RT rule. So since that's true, there is only V left. V is either in that QS block or after because of the S<=V rule. Either way, we know now that this QS block can't be in the 2nd slot, because then there would be nothing available for the first block. So QS is in the first block. Now the Farm can only be the second block, H has to be the third block and the first block is Morningside. Yahtzee! Head to the choices:A. Nope F is secondB. Nope H is thirdC. very possibleD. No S is 1stE. No the 2nd is F and F can only have one

20. This is the first real test of our mettle. Taking a good look at our master diagram, the Qs are the most limited with respect to (H)omestead because they force homestead to be 2nd, since Q can't be third. So for looking for what cannot be true, looking at the most restricted areas is a good start. Of course, the authors know this so they know it's also a good place to mind-eff you. But we take what we are given. A. Has Q and V but a quick diagram shows it's possible. Don't forget to save your diagrams.D. A quick diagram shows it's not possible. -CORRECT ANSWERB. ruled as possible by the diagram for 19, thankfullyC. You can kind of do this in your head: say ST are 2nd and that's homestead. then you can fit RV last and first is Q. PossibleE. It's a possible jumble of C

21. Our worst nightmare, the least restricted factory (morningside) is assigned to Q and V. Luckily you can see two options blossoming, either S will be with these guys or he's before them. diagram those out:

(1) SVQ are first at M -- F is second with R/T -- H is last with R/T(2) S is first at F -- VQ are second at M of course - RT is last at H.

This draws on the numerical dist aspect as well as the s<=v rule. Not rocket science but made harder by the lack of space.

A-D are ruled out by these scenarios.E. remains possible in option 1

22. Another global. This one is actually easy if you quickly browse them to see if any strike you as obvious. Unfortunately I didn't do this but it's not that bad anyway.

Right off the bat E says one of the first two sites has only one exec. Well yeah of course because as we saw in our diagram, F has to be one of those. A. Refer to last question. option one disproves thisB. the correct answer choice in 17 disproves thisC. make a quick hypotheticalD. No, F can be 2nd and that would yield just one Exec. Also, see option one in Q 21.

23. Two quick hypotheticals get you home. if S is in the Farm it's either first or second. Diagram both out

(1) 1st - Farm: S -- 2nd M/H: Q has to be here since it can't be last + /v -- 3rd: H/M RT is here since they are after Q + v/(2) 1st- Morningstar: Q by itself, because it can't be on the farm with somethign. 2nd: the Farm with just S -- 3rd Homestead with VTR.

Attack with that A. could be trueB. Yes always - credited response. if you got here you probably did very well on this LG congrats. if you are like me, you only sniffed at this.C. could be trueD. usually but not alwaysE. only in option 2.